# Challenge Math Challenge - July 2019

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#### nuuskur

Here's what comes to mind
In case of independent random variables $E(XY) = E(X)E(Y)$. By first expanding the expression inside expected value:
\begin{align*} &[(W_u-W_v)^2 + 2(W_u-W_v)W_v + W_v^2]W_v^2 \\ =&(W_u-W_v)^2W_v^2 + 2(W_uW_v -W_v^2)W_v^2 + W_v^4 \\ =&(W_u-W_v)^2W_v^2 + 2W_uW_v^3 - W_v^4 \end{align*}
To get the mgf of Brownian, one computes
$$M_W(x) = \mathbb E(e^{xW_s}) = \mbox{exp}\left ( \frac{1}{2}x^2s\right )$$
Fourth derivative (w.r.t $x$) at $x=0$ gives us the fourth moment, which is (if I calculated correctly) $\mathbb E(W_v^4) = 3v^2$.

Since $W_u-W_v, W_v$ are independent, squaring them preserves independence, so
$$\mathbb E(W_u-W_v)^2W_v^2 = \mathbb E(W_u-W_v)^2\mathbb E(W_v^2) = (u-v)v$$

Not sure what is happening with the middle part, though. Does it vanish?

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#### Math_QED

Science Advisor
Homework Helper
Here's what comes to mind
In case of independent random variables $E(XY) = E(X)E(Y)$. By first expanding the expression inside expected value:
\begin{align*} &[(W_u-W_v)^2 + 2(W_u-W_v)W_v + W_v^2]W_v^2 \\ =&(W_u-W_v)^2W_v^2 + 2(W_uW_v -W_v^2)W_v^2 + W_v^4 \\ =&(W_u-W_v)^2W_v^2 + 2W_uW_v^3 - W_v^4 \end{align*}
To get the mgf of Brownian, one computes
$$M_W(x) = \mathbb E(e^{xW_s}) = \mbox{exp}\left ( \frac{1}{2}x^2s\right )$$
Fourth derivative (w.r.t $x$) at $x=0$ gives us the fourth moment, which is (if I calculated correctly) $\mathbb E(W_v^4) = 3v^2$.

Since $W_u-W_v, W_v$ are independent, squaring them preserves independence, so
$$\mathbb E(W_u-W_v)^2W_v^2 = \mathbb E(W_u-W_v)^2\mathbb E(W_v^2) = (u-v)v$$

Not sure what is happening with the middle part, though. Does it vanish?
You don't have to expand the middle term:

$$E(2(W_u-W_v)W_v^3)=0$$

by independency and the expectation of the third moment here is 0.

#### nuuskur

Hmm, does it hold that $X,Y$ independent implies $X, f(Y)$ independent? I think so, now that I think about it. The sigma algebra generated by $f$ can only get smaller.

#### Math_QED

Science Advisor
Homework Helper
Hmm, does it hold that $X,Y$ independent implies $X, f(Y)$ independent? I think so, now that I think about it. The sigma algebra generated by $f$ can only get smaller.
Yes, it does. If $X,Y$ are independent and $f, g:\mathbb{R} \to \mathbb{R}$ are measurable functions, then $f(X), g(X)$ are independent. This follows from the following simple calculation:

$$P(f(X) \in A, g(Y) \in B) = P(X\in f^{-1}(A), Y\in g^{-1}(B)) = P(X\in f^{-1}(A))P(Y\in g^{-1}(B)) = P(f(X)\in A) P(g(Y)\in B)$$

and $A,B$ are Borel sets.

#### nuuskur

I may have computed something incorrectly. I get that $\mbox{var}(X)$ is negative.

#### Math_QED

Science Advisor
Homework Helper
I may have computed something incorrectly. I get that $\mbox{var}(X)$ is negative.
Well, you didn't include any calculations so I can't see where you made a mistake:

What expression did you get for $E(W_u^2 W_v^2)$? Then you have to calculate a double integral and one has to be careful with the bounds there, but nothing too difficult.

#### nuuskur

I was assuming
$$\mbox{var}(X) = \mathbb E(X^2) - \frac{t^4}{4} = \int_0^t\int _0^t ((u-v)v - 3v^2) dudv - \frac{t^4}{4}$$
Having my doubts about the integrand, most likely incorrectly computed the fourth moment.

#### Math_QED

Science Advisor
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I was assuming
$$\mbox{var}(X) = \mathbb E(X^2) - \frac{t^4}{4} = \int_0^t\int _0^t ((u-v)v - 3v^2) dudv - \frac{t^4}{4}$$
Having my doubts about the integrand, most likely incorrectly computed the fourth moment.
The integrand should contain a minimum! We assumed $u \geq v$ and got an expression. What happens when $u \leq v$ (use symmetry)? Then find the correct expression for $E(W_u^2 W_v^2)$ that should involve $\min\{u,v\}$ somehow.

#### nuuskur

*smacks forehead*
$$u\geq v \implies \mathbb E(W_u-W_v)^2\mathbb E(W_v^2) = (u-v)v,$$
because the second moment of $W_u-W_v$ is its variance. By symmetry it should hold
$$v\geq u \implies \mathbb E(W_u-W_v)^2 = v-u$$
So the inner integral should work out as follows,
$$\int _0^t \mathbb E(W_u^2W_v^2) du = \int _0^v ((v-u)v - 3v^2)du + \int _v^t ((u-v)v -3v^2)du$$

Still not right :/ I am braindead, making some elementary mistake.

#### Math_QED

Science Advisor
Homework Helper
*smacks forehead*
$$u\geq v \implies \mathbb E(W_u-W_v)^2\mathbb E(W_v^2) = (u-v)v,$$
because the second moment of $W_u-W_v$ is its variance. By symmetry it should hold
$$v\geq u \implies \mathbb E(W_u-W_v)^2 = v-u$$
So the inner integral should work out as follows,
$$\int _0^t \mathbb E(W_u^2W_v^2) du = \int _0^v ((v-u)v - 3v^2)du + \int _v^t ((u-v)v -3v^2)du$$

Still not right :/ I am braindead, making some elementary mistake.
I get $E(W_u^2 W_v^2) = 2 \min \{u^2,v^2\} + u v$

#### Math_QED

Science Advisor
Homework Helper
Curious, how you arrive at your expression for $\mathbb E(W_u^2W_v^2)$.
Suppose $u \geq v$. Then
$$E(W_u^2W_v^2) = E[((W_u-W_v) + W_v)^2 W_v^2]$$
$$= E[(W_u-W_v)^2W_v^2] + 2E[(W_u-W_v)W_v^3] + E[W_v^4]$$
$$= E[(W_u-W_v)^2]E[W_v^2] + 2E[W_u-W_v]E[W_v^3] + E[W_v^ 4]$$
$$= (u-v)v + 3v^2 = uv-v^2 + 3v^2 = uv+ 2v^ 2$$
By symmetry:
$$v \geq u \implies E(W_u^2 W_v^2) = uv + 2u^2$$
Conclusion: $$E(W_u^2W_v^2) = uv + 2 \min\{u^2, v^2\}$$
_____________________________

Just like you, I get that $E(X) = t^2/2$.

Hence,
$$Var(X) = EX^2 - (EX)^2 = E[X^2] - t^4/4$$
But unlike you I get
$$E[X^2] = 7t^4/12$$
So the end result should be
$$Var(X) = 7t^4/12 - t^4/4 = 7t^4/12 - 3t^4/12 = 4t^4/12 = t^4/3$$

Did you maybe make a mistake calculating the integral?

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#### Math_QED

Science Advisor
Homework Helper
By symmetry
$$v\geq u \implies \mathbb E(W_u^2W_v^2) = (v-u)v + 3v^2 = 4v^2 -uv.$$
What am I missing?
By symmetry I just mean that we can interchange the roles of $u,v$ in the final expression of the formula. I don't see how you get that. If you don't see it, you can just perform the same calculation that I did but with $u$ and $v$ interchanged and you will get the right answer.

A bit more formally maybe:

I have derived

$$\forall u \geq v \geq 0: E(W_u^2 W_v^2) = uv + 2v^2$$

Now, suppose $v' \geq u'$. The above formula then implies:

$E(W_{u'}^2 W_{v'}^2) = u'v' + 2u'^2$

Now, substitute $u' = u, v' = v$.

#### Math_QED

Science Advisor
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Graah. I am lost again.
$$v\geq u \implies \mathbb E(W_u-W_v)^2 = \mathbb E(W_v-W_u)^2 = v-u,$$
no?
Yes, that's correct. But why is this relevant? Also note my edit of the previous post.

#### nuuskur

Nevermind, now I understand what you are applying symmetry to.

#### Math_QED

Science Advisor
Homework Helper
Nevermind, now I understand what you are applying symmetry to.
Yeah, it can be tricky. I think you learned a lot of this question! Thanks for participating in the harder one!

#### fresh_42

Mentor
2018 Award
Uhmm, guys, what am I missing here?
Eventually, the variance ought to be$$\mbox{var}(X) = \mathbb E(X^2) - \frac{t^4}{4} = \frac{7}{12}t^4$$
But unlike you I get
$$E[X^2]=7t^4/12E[X^2]=7t^4/12$$
P.S.: I just saw that one post has been edited, so maybe this is the reason for this apparently contradicting quotations. If this should have been the case, please mark edits as such, so that the thread will remain readable and contradictions as above can be explained to readers.

#### nuuskur

Oh boy, here we go..
The roots of $x^4-2x^2-2$ are given by $\lvert x\rvert = \sqrt{1\pm\sqrt{3}}$. The polynomial is irreducible.

Put $\alpha := \sqrt{1-\sqrt{3}}, \beta := \sqrt{1+\sqrt{3}}$. Find
$$[\mathbb Q(\alpha,\beta),\mathbb Q] = [\mathbb Q(\alpha,\beta),\mathbb Q(\beta)] [\mathbb Q(\beta),\mathbb Q] = 2[\mathbb Q(\alpha,\beta),\mathbb Q(\beta)]$$
where the degree of $\mathbb Q(\beta) /\mathbb Q$ is clearly $2$, since $1,\beta$ are linearly independent over $\mathbb Q$. It should hold that $1,\alpha$ are a basis of $\mathbb Q(\alpha,\beta)$ over $\mathbb Q(\beta)$, but let's check. Linear independence is clear. Let $k\cdot 1 + l\alpha + m\beta \in\mathbb Q(\alpha,\beta)$, where $k,l,m\in\mathbb Q(\beta)$, then
$$k + l\alpha + m\beta = k + m\beta + l\alpha = (k+m\beta) \cdot 1 + l\alpha$$
where $(k+m\beta)\in\mathbb Q(\beta)$. So this must mean
$$[\mathbb Q(\alpha,\beta),\mathbb Q] = 4.$$
But apparently it doesn't agree with #45.

It remains to study the automorphisms, but I'm not sure of anything at the moment.

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#### Math_QED

Science Advisor
Homework Helper
The polynomial is irreducible.
Why?

Put $\alpha := \sqrt{1-\sqrt{3}}, \beta := \sqrt{1+\sqrt{3}}$. Find
$$[\mathbb Q(\alpha,\beta),\mathbb Q] = [\mathbb Q(\alpha,\beta),\mathbb Q(\beta)] [\mathbb Q(\beta),\mathbb Q] = 2[\mathbb Q(\alpha,\beta),\mathbb Q(\beta)]$$
where the degree of $\mathbb Q(\beta) /\mathbb Q$ is clearly $2$, since $1,\beta$ are linearly independent over $\mathbb Q$.
Linearly independent yes, but generating? How would you write $\beta^2 \in \mathbb{Q}(\beta)$ as a $\mathbb{Q}$-linear combination of $1,\beta?$

The following can be very useful: Consider a field extension $K/F$ with $\alpha \in K$ algebraic over $F$ then

$$[F(\alpha): F] = \deg \left(\min_\mathbb{Q} \alpha\right)$$

#### nuuskur

Wait, are we supposed to compute modulo the polynomial? It's irreducible by Eisenstein (take $p=2$).
So we'd need ${1,\beta,\beta ^2,\beta ^3,\beta ^4}$ to span $\mathbb Q(\beta)$?

#### Math_QED

Science Advisor
Homework Helper
Wait, are we supposed to compute modulo the polynomial? It's irreducible by Eisenstein (take $p=2$).
So we'd need ${1,\beta,\beta ^2,\beta ^3,\beta ^4}$ to span $\mathbb Q(\beta)$?
Yes, irreducible by Eisenstein!

We don't need $\beta^4$. Please look at the statement I wrote down with the minimal polynomial. What is the minimal polynomial of $\beta$ over $\mathbb{Q}$?

#### fresh_42

Mentor
2018 Award
@fresh_42 Is my Q15 proof wrong?
Sorry, no, and, yes, it was correct. It just slipped under the many posts about the Brownian motion and I hadn't seen it.

Nice coordinate system by the way. Pythagoras would have done, too, without complex numbers. Not that others think our high school problems need complex numbers to solve. Pythagoras is also what you called distance formula.

#### fresh_42

Mentor
2018 Award
Oh, I think I understand now, partly..somewhat..hopefully..here we go again
Put $\alpha := \sqrt{1-\sqrt{3}}$ and $\beta := \sqrt{1+\sqrt{3}}$.
By Eisenstein the polynomial is irreducible, thus a minimal polynomial of $\beta$, hence
$$[\mathbb Q(\beta),\mathbb Q] = 4$$
(a basis is given by $1,\beta,\beta ^2,\beta ^3$). Now to find $[\mathbb Q(\alpha,\beta), \mathbb Q(\beta)]$. Note that $\alpha$ is algebraic over $\mathbb Q\subset \mathbb Q(\beta)$ so by relevant hint in #68
$$[\mathbb Q(\alpha,\beta),\mathbb Q(\beta)] = [\mathbb Q(\beta)(\alpha),\mathbb Q(\beta)] = 4,$$
so the field extension $\mathbb Q(\alpha,\beta) /\mathbb Q$ is of degree $16$ by the tower lemma. Not sure how to proceed at the moment. Does one observe how the automorphisms behave on the roots?
How can this be of degree $4$, if $2-\alpha^2=\beta^2\,$?

#### fresh_42

Mentor
2018 Award
Mhh, the hint in #68 is inaccurate, then. I need to check the basics.. again. These side comments are too esoteric.
No, it just means that the minimal polynomial of $\alpha$ over $\mathbb{Q}(\beta)$ is not the same as over $\mathbb{Q}$. You have a basis $\{\,1,\beta,\beta^2,\beta^3\,\}$ of $\mathbb{Q}(\beta)$ over $\mathbb{Q}$ and $\alpha^2 \in \mathbb{Q}(\beta)$. So the only question is, whether $\{\,1,\alpha\,\}$ is linear independent over $\mathbb{Q}(\beta)$, which is a simple linear equation.

#### nuuskur

I understand my mistake: I missed the $\alpha \in K$ requirement. We don't have $\alpha\in\mathbb Q(\beta)$, hence argument that followed is gibberish.
Put $\alpha := \sqrt{1-\sqrt{3}}$ and $\beta := \sqrt{1+\sqrt{3}}$. The polynomial $x^4-2x^2 -2$ is irreducible by Eisenstein, hence minimal polynomial of $\beta$, which yields $[\mathbb Q(\beta),\mathbb Q] = 4$ with a basis of $\{1,\beta,\beta ^2,\beta ^3\}$. Observe that $\alpha^2 = 2-\beta ^2\in\mathbb Q(\beta)$ so it suffices to have $\{1,\alpha\}$ to generate $\mathbb Q(\alpha,\beta) / \mathbb Q(\beta)$ and they are linearly independent over $\mathbb Q(\beta)$. Thus
$$[\mathbb Q(\alpha,\beta),\mathbb Q] = 8.$$
Per definition, the Galois group is the group of automorphisms on $\mathbb Q(\alpha,\beta)$ that pointwise fix $\mathbb Q$. If $\tau$ is such an automorphism and $\alpha$ is a root, then $\tau (\alpha)$ is also a root. An automorphism is therefore determined by how it acts on the roots. This has to be (isomorphic to) a subgroup of $S_4$, one of order $8$ to be exact.

One also notices that the polynomial is even so the roots come in pairs: $\alpha, -\alpha$ and $\beta,-\beta$.

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